0
$\begingroup$

I have got the derivative of a trace w.r.t a real matrix as follows $\frac {\partial}{\partial \mathbf X} tr[\mathbf{(X^TCX)}^{-1}(\mathbf{X^T BX})] = -2 \mathbf{CX(X^TCX)^{-1}X^TBX(X^TCX)^{-1}} + 2\mathbf{BX(X^TCX)^{-1}}$

where $\mathbf {B,C}$ is symmetric.

Now I want to solve the trace w.r.t a complex matrix as follows

$\frac {\partial}{\partial \mathbf X} tr[\mathbf{(X^H CX)}^{-1}(\mathbf{X^H BX})]$

where $\mathbf{B,C}$ conjugate symmertic.

I will feel very gratful if anyone leave me some tips.

$\endgroup$
  • $\begingroup$ I forgot to say that $\mathbf X$ is not a square matrix. $\endgroup$ – freddy Nov 10 '16 at 13:24
  • $\begingroup$ Is it right that you interpret $A:=\partial/(\partial X) tr[\ldots]$ as linear form on the space $\mathbb{R}^{n\times m}\ni X$ (i.e., $A$ applied to a matrix $\delta X\in \mathbb{R}^{n\times m}$ is $\sum_{(i,j)\in \{1,\ldots,n\}\times\{1,\ldots,m\}} A_{i,j} \cdot \delta X_{i,j}$)? $\endgroup$ – Tobias Nov 10 '16 at 13:42
0
$\begingroup$

Maybe not the final solution but a step forward:

Following calculation is valid only under the assumption that $X$ is square and regular.

You can simplify $(X^H CX)^{-1}X^H BX = X^{-1}C^{-1}X^{-H} X^H B X = X^{-1}C^{-1}B X$.

The Gateaux-derivative $\delta A:=\left.\partial_\varepsilon A(X+\varepsilon \delta X)\right|_{\varepsilon=0}$ of the inverse-matrix operation $A(X):=X^{-1}$ results from

\begin{align*} 1 &= A(X) X,\\ 0 &= \delta(AX) = \delta A\, X + A\,\delta X,\\ \delta A &= -A\,\delta X\,X^{-1}\\ &= \underline{\underline{X^{-1}\,\delta X\,X^{-1}}}. \end{align*}

This helps with the calculation of the Gateaux-derivative of your original equation: \begin{align*} \delta \operatorname{tr}[X^{-1}C^{-1}B X] &= \operatorname{tr}[X^{-1} \delta X X^{-1} C^{-1}B X + X^{-1}C^{-1}B \delta X] \end{align*}

$\endgroup$
  • $\begingroup$ I forgot to say that $\mathbf X$ is not a square matrix. $\endgroup$ – freddy Nov 10 '16 at 13:24
  • $\begingroup$ @freddy The general case where $X$ is not square can be treated analogously. It is just more work. $\delta( X^H ) = (\delta X)^H$. $\endgroup$ – Tobias Nov 10 '16 at 14:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.